Given is a 2D scenario of a particle within a container: enter image description here

The circumference shows all the possible 2D spatial directions for a given momentum value of the particle. One of those momentum directions is $p_e$.

From what I understand, the number of states within that $p_e$ is given by: $$n_e = \sqrt{\frac{L_x \cdot p_{e,x}}{h}^2 + \frac{L_y \cdot p_{e,y,}}{h}^2}$$ Where $p_{e,x}$ and $p_{e,y}$ are the projections of $p_e$ in the $x$ and $y$ coordinates.

Since $p_e$ has a spatial direction, this means that the container has a length in that same spatial direction. Let's call that container length in that same spatial direction $L_e$, illustrated as follows:

enter image description here

Does this $L_e$ have any relationship with the number of states in momentum $p_e$? For example, is the following equation valid? $$n_e = \sqrt{\frac{L_x \cdot p_{e,x}}{h}^2 + \frac{L_y \cdot p_{e,y,}}{h}^2} = \frac{L_e p_e}{h}$$

EDIT: I forgot to state to use periodic boundary condition in such a case, since rigid boundaries gives zero momentum to particles. So my question is: is it possible to calculate the number of states in a single momentum that has just one direction in p-space? If so, how is this derived from the formula: $$n=\frac{V \cdot p^3}{h^3}$$

  • $\begingroup$ Don't all eigenstates of a particle in a box have zero momentum? $\endgroup$ – Ivan Neretin Nov 6 '18 at 15:02
  • $\begingroup$ Apologies, I should have stated to use periodic boundary conditions $\endgroup$ – JohnnyGui Nov 14 '18 at 21:56

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